If the displacement of a particle is `x=t^(3)-4t^(2)-5t`, then the velocity of particle at t=2 is

Options

A. `-9" units/sec"`
B. 9 units/sec
C. `-18" units/sec"`
D. 18 units/sec

Answer

To find the velocity of the particle at time \( t = 2 \), we need to follow these steps:

### Step 1: Write down the displacement function
The displacement of the particle is given by:
\[ x(t) = t^3 - 4t^2 - 5t \]

### Step 2: Differentiate the displacement function to find the velocity function
The velocity \( v(t) \) is the derivative of the displacement \( x(t) \) with respect to time \( t \):
\[ v(t) = \frac{dx}{dt} = \frac{d}{dt}(t^3 - 4t^2 - 5t) \]

Using the power rule of differentiation:
\[ v(t) = 3t^2 - 8t - 5 \]

### Step 3: Substitute \( t = 2 \) into the velocity function
Now we need to find the velocity at \( t = 2 \):
\[ v(2) = 3(2^2) - 8(2) - 5 \]

Calculating each term:
\[ = 3(4) - 8(2) - 5 \]
\[ = 12 - 16 - 5 \]
\[ = 12 - 21 \]
\[ = -9 \]

### Step 4: State the final answer
Thus, the velocity of the particle at \( t = 2 \) is:
\[ v(2) = -9 \text{ units per second} \]